Electronic Journal of Differential Equations, Vol. 1998(1998), No. 11, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
ftp (login: ftp) 147.26.103.110 or 129.120.3.113
BARRIERS ON CONES FOR DEGENERATE
QUASILINEAR ELLIPTIC OPERATORS
Michail Borsuk
Dmitriy Portnyagin
Abstract. Barrier functions w = |x|λ Φ(ω) are constructed for the first boundary
value problem as well as for the mixed boundary value problem for quasilinear elliptic
second order equation of divergent form with triple degeneracy on the n-dimensional
convex circular cone:
d
(|x|τ |u|q |∇u|m−2 uxi ) = µ|x|τ |u|q−1 sgn u|∇u|m ,
dxi
where −1 < µ ≤ 0, q ≥ 0, m > 1, τ > m − n.
Introduction
Lately many mathematicians have been considering nonlinear problems for elliptic degenerate equations; see e.g. [1] and its extensive bibliography. In the present
paper, we take a first step on the investigation of the behaviour of solutions of
boundary value problems for quasilinear elliptic second-order equations with triple
degeneracy. We study the problem
d
Lu ≡
(|x|τ |u|q |∇u|m−2 uxi ) = µ|x|τ |u|q−1 sgn u|∇u|m , x ∈ G0 ,
dxi
(1)
−1 < µ ≤ 0, q ≥ 0, m > 1, τ > m − n ,
where G0 is an n-dimensional convex circular cone with its vertex at the origin O,
having Γ0 as lateral area; and Ω is a domain, on the unit sphere, with a smooth
boundary ∂Ω. We shall construct functions playing a fundamental role in the
study of the behaviour of solutions to elliptic boundary value problems in the
neighbourhood of the irregular boundary point; see e.g. [2–6]. The special structure
of the solution near a conical point is of particular interest in physical applications,
[7–9]. It is also used for improving numerical algorithms, [10–12].
The proof of the estimates for the solution is based on the observation that the
function r λ Φ(ω) is usable as barrier in the above problem. By the weak comparison
principle in [2, chapt. 10], it is possible to verify that the assumptions of this
principle are fulfilled. Since (1) is equivalent to
d |∇u|m−2 uxi + τ |x|−2 |∇u|m−2 (x∇u) + (q − µ)|u|−1 sgn u|∇u|m = 0,
dxi
x ∈ G0 , −1 < µ ≤ 0, q ≥ 0, m > 1, τ > m − n
1991 Mathematics Subject Classifications: 35J65, 35J70, 35B05, 35B45, 35B65.
Key words and phrases: quasilinear elliptic equations, barrier functions, conical points.
c
1998
Southwest Texas State University and University of North Texas.
Submitted May 15, 1997. Published April 17, 1998.
1
Michail Borsuk & Dmitriy Portnyagin
2
EJDE–1998/11
on the set where u 6= 0, one can obtain the bound for solution near conical boundary
point. In this setting, finding of exact value of the exponent λ is very important
and very difficult. For the case of a planar bounded domain with corner boundary
points, the exact value of the exponent λ will be calculated explicitly.
Let us transfer the above problem to spherical coordinates with the pole at the
point O.
x1 = r cos ω1 ,
x2 = r cos ω2 sin ω1 ,
..
.
xn−1 = r cos ωn−1 sin ωn−2 . . . sin ω1 ,
xn = r sin ωn−1 . . . sin ω1 ,
where r = |x| > 0, 0 < ωi < π for i = 1, . . . , n − 2, and 0 < ωn−1 < 2π.
The differential operator L takes the form
n
1X d
τ
q
m−2 J ∂u
Lu =
r |u| |∇u|
,
J i=1 dξi
Hi2 ∂ξi
where J = r n−1 sinn−2 ω1 , . . . , sin ωn−2 ; H1 = 1; ξ1 = r; ξi+1 = ωi and Hi+1 =
√
r qi , for i = {1, n − 1}; q1 = 1; qi = (sin ω1 . . . sin ωi−1 )2 for i = {2, n − 1}.
We shall seek the solution of (1) as a function of the form u = r λ Φ(ω) with
Φ(ω) ≥ 0. Then Φ(ω) satisfies the equation
n−1
1 X d
j(ω) 2 2
∂Φ
(λ Φ + |∇ω Φ|2 )(m−2)/2 |Φ|q
j(ω)
dωk
qk
∂ωk
k=1
+ λ[λ(q + m − 1) + τ + n − m](λ2 Φ2 + |∇ω Φ|2 )(m−2)/2 Φ|Φ|q
= µΦ|Φ|q−2 (λ2 Φ2 + |∇ω Φ|2 )m/2 , ω ∈ Ω,
where |∇ω Φ|2 =
Pn−1
1 ∂Φ 2
j=1 qj ( ∂ωj ) ,
(2)
and j(ω) = sinn−2 ω1 . . . sin ωn−2 .
The Dirichlet problem
First we consider the Dirichlet problem for (1) when u|Γ0 = 0. On multiplying
(2) by Φ(ω) and integrating by parts over Ω we obtain
Z
(m−2)/2
λ2 Φ2 + |∇ω Φ|2
|Φ|q |∇ω Φ|2 dΩ
Ω
Z
= λ[λ(q + m − 1) + τ + n − m]
Z
−µ
Z
≡
Ω
λ2 Φ2 + |∇ω Φ|2
(m−2)/2
|Φ|q+2 dΩ
Ω
|Φ|q
m/2
λ2 Φ2 + |∇ω Φ|2
dΩ
Ω
λ2 Φ2 + |∇ω Φ|2
(m−2)/2
|Φ|q ×
λ[λ(q + m − 1) + τ + n − m]Φ2 − µ(λ2 Φ2 + |∇ω Φ|2 ) dΩ .
EJDE–1998/11
Barriers on cones
3
Hence, it follows that
Z
(m−2)/2 q
(1 + µ)
λ2 Φ2 + |∇ω Φ|2
|Φ| |∇ω Φ|2 dΩ
Ω
Z
= λ[λ(q + m − 1 − µ) + τ + n − m]
λ2 Φ2 + |∇ω Φ|2
(m−2)/2
|Φ|q+2 dΩ .
Ω
Since Φ(ω) 6≡ 0 and µ > −1, we have
λ[λ(q + m − 1 − µ) + τ + n − m] > 0 .
(*)
We shall consider the case of Φ(ω) not depending on ω2 , . . . , ωn−1 ; so that Φ is
a function of a single angular coordinate ω1 = ω ∈ (−ω0 /2, ω0 /2), 0 < ω0 < π.
Such function Φ(ω) satisfies the boundary value problem for ordinary differential
equation
2
2
[(m − 1)Φ0 + λ2 Φ2 ]ΦΦ00 + (λ2 Φ2 + Φ0 )×
n
o
2
(q − µ)Φ0 + λ[λ(q + m − 1 − µ) + τ + n − m]Φ2 + (n − 2)ΦΦ0 cot ω
2
2
+ (m − 2)λ Φ Φ
02
= 0,
(ODE)
ω ∈ (−ω0 /2, ω0 /2)
Φ(−ω0 /2) = Φ(ω0 /2) = 0 .
By making the substitution y = Φ0 /Φ and y 0 + y 2 = Φ00 /Φ, we arrive to
[(m − 1)y 2 + λ2 ]y 0 + (m − 1 + q − µ)(y 2 + λ2 )2
+ [λ(τ + n − m) + (n − 2)y cot ≤](y 2 + λ2 ) = 0;
ω ∈ (−
ω0 ω0
, ). (3)
2 2
Let us now verify that
Φ(−ω) = Φ(ω),
y 0 (−ω) = y 0 (ω),
ω0 ω0
∀ω ∈ (− , ) .
2 2
y(−ω) = −y(ω),
Φ0 (−ω) = −Φ0 (ω),
Putting ω = 0 we obtain y(0) = 0. Therefore, it is sufficient to consider the
equation only on the interval (0, ω0 /2). Since cot ω > 0 on (0, ω0 /2), from (3) and
(*) it follows that
[(m − 1)y 2 + λ2 ]y 0 + (n − 2)y(y 2 + λ2 ) cot ω < 0 ,
ω ∈ (0,
ω0
).
2
ω ∈ (0,
ω0
);
2
Let us solve the Cauchy problem
[(m − 1)y 2 + λ2 ]y 0 + (n − 2)y(y 2 + λ2 ) cot ω = 0,
y(0) = 0 .
We obtain
Z
(m − 1)y 2 + λ2
dy = −(n − 2)
y(y 2 + λ2 )
Z
cot ωdω + const
(4)
Michail Borsuk & Dmitriy Portnyagin
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EJDE–1998/11
which implies
y(y 2 + λ2 )(m−2)/2 = Csin
2−n
ω
y(0) = 0 .
This, in turn, implies C = 0 and y ≡ 0.
Comparing the solution of (4) with that of the Cauchy problem, we deduce that
y(ω) ≤ 0. Since cot ω > 0 and y ≤ 0 on our interval, by (3) we have
[(m − 1)y 2 + λ2 ]y 0 + [(m − 1 + q − µ)(λ2 + y 2 ) + λ(τ + n − m)](λ2 + y 2 )
= −(n − 2)y(y 2 + λ2 ) cot ω ≥ 0,
ω ∈ (0,
ω0
).
2
Thus, for (0, ω0 /2) we have,
[(m − 1)y 2 + λ2 ]y 0 ≥ −[(m − 1 + q − µ)(λ2 + y 2 ) + λ(τ + n − m)](λ2 + y 2 )
y(0) = 0 .
Similarly by the comparison theorem, we obtain y(ω) ≥ z(ω), where z(ω) with
ω ∈ (0, ω0 /2) is the solution to Cauchy problem
[(m − 1)z 2 + λ2 ]z 0 = −[(m − 1 + q − µ)(λ2 + z 2 ) + λ(τ + n − m)](λ2 + z 2 ),
z(0) = 0 .
On solving the latter, we obtain the expression for z in the implicit form
m−1
m−1+q−µ
q
λ2 +
m−2
+ λ τ +n−m
τ +n−m
λ m−1+q−µ
arctan q
z
τ +n−m
λ2 + λ m−1+q−µ
(5)
m−2
z
+ω +
arctan( ) = 0 .
m−n−τ
λ
By combining the obtained results, we conclude that
0 ≥ y(ω) ≥ z(ω) .
(6)
Let us now return to the equation for y(ω). On making the substitution ϕ = ln Φ,
w(ϕ) = y 2 (ϕ),
dω
w0 (ϕ) = 2yy 0 (ϕ) = 2y y 0 (ω) = 2y 0 (ω) ,
dϕ
we obtain
1
[(m − 1)w + λ2 ]w0 + [(m − 1 + q − µ)(λ2 + w) + λ(τ + n − m)](λ2 + w)
2
√
−(n − 2) w(w + λ2 ) cot ω = 0 ,
√
where we have used y = ± w and y < 0. As we did above, we obtain a differential
inequality for w,
1
[(m − 1)w + λ2 ]w0 + [(m − 1 + q − µ)(λ2 + w) + λ(τ + n − m)](λ2 + w) > 0 .
2
EJDE–1998/11
Barriers on cones
5
Integrating the respective differential equation
1
[(m − 1)w + λ2 ]w 0 + [(m − 1 + q − µ)(λ2 + w) + λ(τ + n − m)](λ2 + w) = 0
2
we obtain
m−2
ln(λ2 + w)
m−n−τ
m−1
m−2
+
+λ
ln((m − 1 + q − µ)(λ2 + w) + λ(τ + n − m))
m−1+q−µ
τ +n−m
+ 2 ln Φ = ln C .
λ
Solving the latter expression with the respect to Φ we obtain
Φ2 (ω) = C 2
(m − 1 + q − µ)(λ2 + w) + λ(τ + n − m)
λ2 + w
λ(m−2)/(m−n−τ )
×
[(m − 1 + q − µ)(λ2 + w) + λ(τ + n − m)]−(m−1)/(m−1+q−µ) .
Now it is evident that w = z 2 (ϕ) and w = y 2 (ϕ). From (6) and w ≤ w, it follows
that
λ(m−2) − m−1
1−m
λ(τ + n − m) m−n−τ m−1+q−µ
2
2 2
2 m−1+q−µ
Φ (ω) = C (z + λ )
m−1+q−µ+
.
(z 2 + λ2 )
Whence it follows that
(m−1)
Φ(ω) ∼ |z|− m−1+q−µ
as
|z| → +∞ .
Since y 2 ≤ z 2 , it follows that 1/z 2 ≤ 1/y 2 , and it is now clear that
lim
ω→(ω0 /2)−0
z(ω) = −∞
0
(since Φ(ω0 /2) = 0). Furthermore, since y = ΦΦ < 0 and Φ > 0 on (0, ω0 /2), Φ0 < 0.
i.e., Φ(ω) decreases on (0, ω0 /2) from the positive value Φ(0) to Φ(ω0 /2) = 0. Φ
does not vanish anywhere else in (0, ω0 /2), otherwise it should increase somewhere.
From this equation we have
y 0 = −[(m − 1 + q − µ)(y 2 + λ2 ) + λ(τ + n − m)]
−(n − 2)y
y 2 + λ2
cot ω → −∞ as
(m − 1)y 2 + λ2
(y 2 + λ2 )
(m − 1)y 2 + λ2
y → −∞ .
That is to say y(ω) decreases in the vicinity of the point −ω, when y → −∞. It
is possible only at −ω = ω0 /2, (when passing ω → ω20 − 0). On performing the
passage to the limit ω → ω20 − 0 in (5), and taking into account that z → −∞, we
obtain
ω0
m−2
+
π
τ +n−m
−1/2
m−1
m−2
λ[λ(m − 1 + q − µ) + τ + n − m]
=
.
+λ
m−1+q−µ
τ +n−m
m−1+q−µ
Michail Borsuk & Dmitriy Portnyagin
6
EJDE–1998/11
Hence we obtain an explicit expression for λ,
λ=
(
m(m − 2) − 2(m − 2)t − t2
+
t + 2(m − 2)
p
π
×
2ω0 (m − 1 + q − µ)
[t2 + 2(m − 2)t + m2 ][t2 + 2(m − 2)t + (m − 2)2 ]
t + 2(m − 2)
(7)
where t = (τ + n − m)ω0 /π.
In the case of n = 2, τ = 0 we obtain
p
(π − ω0 )[m(π − ω0 ) + (m − 2)2 (π − ω0 )2 + 4(m − 1)π 2 ]
(m − 1)
λ=
+
.
(m − 1 − µ + q)
2ω0 (2π − ω0 )(m − 1 − µ + q)
(8)
In the case of τ = µ = q = 0, n = 2 we get the result of [13]. If m = n = 2, from
(7) we get
r 2
2π
+ τ2 − τ
ω0
λ=
.
(9)
2(1 + q − µ)
For the case n = 3, we assume that τ = 0. We shall seek a solution of the form
u = r λ Φ(ω) sinλ Φ, with Φ ∈ (0, π) and ω ∈ (−ω0 /2, ω0 /2). Then we obtain for
Φ(ω) a problem which coincides with (ODE) with n = 2, and so for λ we have the
Expression (8).
The mixed boundary value problem
Now we consider the mixed boundary value problem in the planar domain G0 =
{(r, ω) | r > 0, 0 < ω < ω0 < π}, with a corner boundary point,
d q
m−2
q−1
m
|u| |∇u|
uxi = µ|u| sgn u|∇u| ,
dxi
∂u u ω=ω = 0 ,
= 0,
0
∂x2 ω=0
x ∈ G0 ,
where ω0 is the angle with the vertex at the point O. By a process analogous to
the one above, we come to the expression
λ=
(m − 1)
(m − 1 − µ + q)
p
(π − 2ω0 )[m(π − 2ω0 ) + (m − 2)2 (π − 2ω0 )2 + 4(m − 1)π 2 ]
+
. (10)
8ω0 (π − ω0 )(m − 1 − µ + q)
Obviously, this expression coincides with (8) for the Dirichlet problem, if in the
latter we put 2ω0 instead of ω0 .
Therefore, barrier functions w = r λ Φ(ω) have been constructed for the first
boundary value problem for the equation (1), and for the mixed boundary value
problem for (1) with τ = 0.
)
,
EJDE–1998/11
Barriers on cones
7
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Michail Borsuk
Department of Applied Mathematics
Olsztyn University of Agriculture and Technology
10-957 Olsztyn-Kortowo, Poland
E-mail address: borsuk@art.olsztyn.pl
Dmitriy Portnyagin
Department of Physics
Lvov State University
290602 Lvov, Ukraine